# In derivation of capillary rise we take the upwards pressure as $2T/r$. How?

In the derivation of capillary rise, we take excess pressure due to surface tension in upward direction as equal to $$2T/r$$.

• Refer to Young-Laplace equation. The answers given are correct but in think this is more fundamental – Ishan Jawale Jun 15 '19 at 13:05

The derivation can be thought of as this:

Let,

• Radius of Capillary be $$r$$

• Density of the liquid $$\rho$$

• Height of the liquid be $$h$$

• Surface Tension of Liquid be $$T$$

• Contact angle $$\theta$$

Weight of liquid inside capillary = Volume * Density * $$g$$ $$=\pi r^2 h \rho g$$Which is the downward force, and the force that is balancing this is the force due to Surface tension.

Now, Surface tension is defined as the Force acting on a line which is on the surface. In this case the surface tension is acting on the circumference.

Hence, total force upwards: component of Surface Tension upwards * length of the line it acts on $$T\cos\theta (2\pi r)$$ The $$\sin\theta$$ components gets cancelled as it is radially outward throughout the circumference.

Equating the forces we get: $$\pi r^2 h \rho g = T\cos\theta (2\pi r)$$ $$\implies h = \frac{2T\cos\theta}{r\rho g}$$

Note: In cases of some liquids the $$\theta$$ is very close to $$0$$ degrees and hence the cos$$\theta$$ term can be taken as $$1$$.

We know that $$p=dgh$$, where $$d=$$ density, $$g=$$ acceleration due to gravity, $$h=$$ height. Therefore, from the above equation $$p=2T/r$$ where $$\cos \theta =1$$.